<OU1 60443 >m OSMANIA UNIVERSITY LIBRARY Call No. S 7 *' &/DS$> Accession No. This book should be returned on or^oefore the date last marked below. SPHERICAL TRIGONOMETRY AFTER THE CESARO METHOD SPHERICAL TRIGONOMETRY AFTER THE CESARO METHOD J. D. H. DONNAY. E.M., Ph.D. 1945 INTERSCIENCE PUBLISHERS. INC. - NEW YORK. N. Y. Copyright, 1945, by INTERSCIENCE PUBLISHERS, INC. 215 Fourth Avenue, New York 3, N. Y. Printed in the United State* of America by the Lancaster Press, Lancaster, Pa. To THE MEMORY OF GIUSEPPE CESARO 1849 - 1939 CRT8TALLOGRAPHER AND MINERALOGIST PROFESSOR AT THE UNIVERSITY OF LIEGE PREFACE From a practical standpoint spherical trigonometry is useful to engineers and geologists, who have to deal with surveying, geodesy, and astronomy; to physicists, chemists, mineralogists, and metallurgists, in their common study of crystallography; to Navy and Aviation officers, in the solu- tion of navigation problems. For some reason, however, spherical trigonometry is not recognized as a regular subject in many American college curricula. As a consequence, the teacher of a science for which a working knowledge of spher- ical triangles is desirable usually finds he has to impart it to the students himself, or to use ready-made formulae that his listeners have never seen, let alone derived, before. This book has been written in an attempt to meet this situation. It can be covered in about ten to twelve lectures and could well serve as a text in a one-unit course for a quarter or a semester. It aims at giving the strict minimum, as briefly as possible. The straightforward and time-saving Cesaro method seems particularly suitable for this purpose, tying together as it does, from the outset, the concepts of spherical and plane trigonometry. This method has with- stood the test of experience. For years Belgian students have thrived on it. Personally I have taught it for eight years to crystallography students at the Johns Hopkins University, and for two years to freshmen classes at Laval University, with gratifying results. The order in which the subject matter is arranged may appear unorthodox. It has proved satisfactory, however, from the teaching point of view. Through the use of the stereographic projection (Ch. I), the concept of spherical excess somewhat of a stumbling block is mastered from vii viii PREFACE the start. A working knowledge of Cesaro's key-triangles is acquired, as soon as they are established (Ch. II), through the derivation of Napier's and Delambre's formulae and several expressions of the spherical excess (Ch. III). This much insures the understanding of the method. The treat- ment of the oblique-angled triangle (Ch. IV) is thereby so simplified that one can dispense with that of the right-angled triangle. The latter properly follows as a particular case (Ch. V). Examples of computations are given (Ch. VI), both with logarithms and with the calculating machine. They are followed by a selection of problems, completely worked out (Ch. VII), and a number of exercises with answers (Ch. VIII). Cross-references are made in the theoretical chapters to appropriate applications. Proofs believed to be new are marked by asterisks. In conclusion may I be permitted to say that this booklet was planned jointly by Cesaro and myself, several years ago. He who was to have been the senior author passed away shortly afterwards. In introducing Cesaro's elegant method to the English-speaking public, I would like to think that my writing will reflect his influence and, to some degree at least, his reverence for simplicity and rigor. J. D. H. D. Hercules Powder Company Experiment Station Wilmington, Delaware April 1945 CONTENTS PAGE Dedication v Preface . vn Introduction 3 1. Purpose of Spherical Trigonometry 3 2. The Spherical Triangle 4 3. Comparison between Plane and Spherical Trigo- nometry 5 4. Fields of Usefulness of Spherical Trigonometry. 5 I. The Stereographic Projection 7 1. Definition 7 2. First Property of the Stereographic Projection. 7 3. Second Property of the Stereographic Projection. 10 II. Cesaro's Key-Triangles 13 1. Cesaro's "Triangle of Elements" 13 2. Cesaro's "Derived Triangle" 16 3. Polar Triangles 18 4. Derivation of the Formulae of Spherical Trigo- nometry 20 Exercises 21 III. How the Key-Triangles are Put to Work 22 1. Napier's Analogies 22 2. Delambre's Formulae 22 3. Euler's Formula 23 4. Lhuilier's Formula 24 5. Cagnoli's Formula 25 iz CONTENTS PAGE 6. The Spherical Excess in Terms of Two Sides and Their Included Angle 26 Exercises 26 IV. Relations between Four Parts of a Spherical Tri- angle 28 1. Relation between the Three Sides and One Angle 28 2. Expression of the Half-Angles in Terms of the Three Sides 29 3. Relation between Two Sides and Their Opposite Angles 29 4. Relation between the Three Angles and One Side 30 5. Expression of the Half-Sides in Terms of the Three Angles 30 6. Relation between Two Sides, their Included Angle, and the Angle Opposite One of them .... 32 Exercises 34 V. Right-Angled Triangles 36 1. The Right-Angled Triangle, a Special Case of the Oblique-Angled Triangle 36 2. Relations between the Hypotenuse and Two Sides or Two Angles (the product formulae) .... 37 3. Relations between the Hypotenuse, One Side, and One Angle (the ratios of sines and tangents) . 37 4. Relations between Two Angles and One Side, or between Two Sides and One Angle (the ratios of cosines and cotangents) 38 5. Napier's Rule 39 VI. Examples of Calculations 41 1. Solution of Triangles 41 2. Formulae Adapted to Logarithmic Computation 42 3. Numerical Calculations 44 4. First Example: given a, 6, c, solve for A 45 5. Second Example: given A, B, C, solve for a, 6, c. 48 CONTENTS zi PAGE 6. Third Example: given a, 6, and C = 90, solve for A 50 7. Fourth Example: spherical distance between two points given by their geographical coordinates. . 51 8. Numerical Application 52 VII. Problems (Selected Problems With Complete Solu- tions) 55 VIII. Exercises 70 Answers to Exercises 74 Appendix 1. Spherical Areas 76 Appendix 2. Formulae of Plane Trigonometry 78 Index 81 SPHERICAL TRIGONOMETRY AFTER THE CESARO METHOD INTRODUCTION 1. Purpose of Spherical Trigonometry. Spherical trigonome- try is essentially concerned with the study of angular rela- tionships that exist, in space, between planes and straight lines intersecting in a common point O. A bundle of planes passed through intersect one another in a sheaf of straight lines. Two kinds of angles need therefore be considered: angles between lines l and angles between planes (dihedral angles). The spatial angular relationships are more easily visualized on a sphere drawn around O with an arbitrary radius. Any line through is a diameter, and any plane through a diametral plane, of such a sphere. The former punctures the sphere in two diametrically opposite points, the latter inter- sects it along a great circle. The angle between two lines is measured on the sphere by an arc of the great circle whose plane is that of the two given lines. The angle between two planes is represented by the angle between the two great circles along which the given planes intersect the sphere. Indeed, by definition, the angle between the great circles is equal to the angle between the tangents to the circles at their point of intersection, but these tangents are both perpen- dicular to the line of intersection of the two given planes, hence the angle between the tangents is the true dihedral angle. An open pyramid, that is to say a pyramid without base, whose apex is made the center of a sphere determines a spherical polygon on the sphere. The vertices of the polygon are the points where the edges of the pyramid puncture the 1 From now on, the word line will be used to designate a straight line, unless otherwise stated. 4 INTRODUCTION sphere; the sides of the polygon are arcs of the great circles along which the faces of the pyramid intersect the sphere. The angles of the polygon are equal to the dihedral angles between adjacent faces of the pyramid. The sides of the polygon are arcs that measure the angles of the faces at the apex of the pyramid, that is to say, angles between adjacent edges. A trihedron is an open pyramid with three faces. The three axes of co-ordinates in solid analytical geometry, for instance, are the edges of a trihedron, while the three axial planes are its faces. Consider a trihedron with its apex at the center of the sphere. It determines a spherical triangle on the sphere. 2 In the general case of an oblique trihedron, an oblique-angled spherical triangle is obtained, that is to say, one in which neither any angle nor any side is equal to 90. The main object of spherical trigonometry is to investi- gate the relations between the six parts of the spherical tri- angle, namely its three sides and its three angles. 2. The Spherical Triangle. The sides of a spherical triangle are arcs of great circles. They can be expressed in angular units, radians or degrees, since all great circles have the same radius, equal to that of the sphere. As a consequence of the conventional construction by means of which the spherical triangle has been defined (Sn. 1), any side must be smaller than a semi-circle and, likewise, any angle must be less than 180. By considering the trihedron whose apex is at the center of the sphere, we see that the sum of any two sides of a spherical triangle is greater than the third side, that any side is greater than the difference between the other two sides, and that the sum of all three sides (called the perimeter) is less than 360. Because any angle of a spherical triangle is less than 180, the sum of all three angles is obviously less than 540. It is greater than 180, as we shall see later (Ch. II, Sn. 1). This is the "Eulerian" spherical triangle, the only one to be considered in this book. USEFULNESS OF SPHERICAL TRIGONOMETRY 5 3. Comparison between Plane and Spherical Trigonometry. In plane trigonometry, you draw triangles in a plane. Their sides are segments of straight lines. The shortest path from one point to another is the straight line that connects them. The distance between two points is measured, along the straight line, in units of length. The sum of the angles of a plane triangle is 180. Through any point in the plane, a straight line can be drawn parallel to a given line in the plane (Euclidian geometry). In spherical trigonometry, triangles are drawn on a sphere. Their sides are arcs of great circles. The shortest path (on the sphere) from one point to another is the great circle that connects them. The (spherical, or angular) distance between two points is measured, along the great circle, in units of angle. The sum of the angles of a spherical triangle is greater than 180. Through a point on the sphere, no great circle can be drawn parallel to a given great circle on the sphere (spherical geometry is non-Euclidian). 4. Fields of Usefulness of Spherical Trigonometry. Most problems dealing with solid angles can be reduced to questions of spherical trigonometry. Such problems crop up in the study of geometrical polyhedra. Problems of solid analytical geometry in which planes and lines pass through the origin usually have trigonometric solutions. Problems involving spatial directions around one point are encountered in crystallography. Well formed crystals are bounded by plane faces. From a point taken anywhere inside the crystal, drop a perpendicular on each face; this face normal defines the direction of the face. Relationships between the inclinations of the faces relative to one another appear in the network of spherical triangles which the sheaf of face normals determines on a sphere drawn around 0. Surveying is concerned with such small regions of the earth surface that they can be considered plane in a first approxima- tion. Geodesy deals with larger regions, for which the curva- 6 INTRODUCTION ture of the earth must be taken into account. In a second approximation the earth is taken as spherical, and the for- mulae of spherical trigonometry are applicable. (Further re- finements introduce corrections for the lack of perfect sphe- ricity of the "geoid.") In astronomy the application of spherical trigonometry is obvious. The observer occupies a point that is very nearly the center of the celestial sphere around the earth. Each line of sight is a radius of the sphere. To the observer who is not aware of, or concerned with, the distances from the earth to the heavenly bodies, the latter appear to move on a sphere. The angle subtended by two stars, as seen by the observer, will thus become a side in a spherical triangle. Navigation techniques, either on the high seas or in the air, being based on astronomical observations, likewise depend on the solution of spherical triangles. CHAPTER I THE STEREOGRAPHIC PROJECTION 1. Definition. The problem of representing a sphere on a plane is essentially that of map projections. Of the many types of projections that have been devised, one of the most ancient is the stereographic. In geographical parlance, used for convenience, the pro- jection plane is the plane of the equator, and the projection point, the South Pole. Points in the Northern Hemisphere are projected inside the equatorial circle; points in the South- ern Hemisphere, outside the equatorial circle; any point on the equator is itself its own stereographic projection (Fig. 1). The North Pole is projected in the center of the projection; the South Pole, at infinity. FIG. 1. The stereographic projection. 2. First Property of the Stereographic Projection. Circles are projected as circles or straight lines. If the circle to be projected passes through S, its projection is a straight line. This is obvious, since the projection of the circle is the inter- 7 8 I. THE STEREOGRAPHIC PROJECTION section of two planes: the plane of the circle and the plane of the equator. Note that if the given circle is a great circle passing through S (hence, a meridian) its projection is a diameter of the equator. ' If the circle to be projected does not pass through S, its projection is a circle. The proof 1 of this is based on the following theorem. Consider (Fig. 2) an oblique cone with vertex S and circular base AB. Let the plane of the drawing be a section through S and a diameter AB of the base. The circular base is pro- jected on the drawing as a straight line AB. A section ab, perpendicular to the plane of the drawing, and such that the angles SAB, S6a are equal, is called sub-contrary (or anti- parallel) to the base. Theorem: In an oblique cone having a circular base, any section sub-contrary to the base is circular. FIG. 2. Sub-contrary sections. Take a section cd parallel to the base and, hence, obviously circular. The two sections ab and cd intersect along a com- mon chord, projected at a point n, which bisects the chord. Let x be the length of the semi-chord. The triangles can and bdn are similar (angles equal each to each), hence an : nd 1 This proof can be omitted in a first reading. FIRST PROPERTY 9 = cn : nb, or an.nb = cn.nd. Because cd is circular, cn.nd = x 2 . Hence an . nb = x 2 , and ab is also circular. 2 Now consider a section of the sphere of projection cut perpendicular to the intersection of the equatorial plane EE' and the plane of the circle AB to be projected (Fig. 3). The angles SAB, S6a (marked on the drawing) are equal, since the measure of SAB = KSE' + BE 7 ) and that of S6a = |(SE + BE') are equal (because SE = SE' = 90). The sections ab and AB of the cone of projection are therefore sub-contrary. Since AB is circular, so is ab. S FIG. 3. Projected circle, a circle. Remark. The center of the projected circle is the projec- tion of the vertex of the right cone tangent to the sphere along the given circle. Let C be the vertex of the right cone tangent to the sphere along the given circle AB (Fig. 4). Join CS, intersecting the equator in c and the sphere in D. The angles marked a are equal as having the same measure (one half arc AD). Likewise for the angles marked 0, 7, d. The Law of Sines, applied to the triangles Sac and Sc6, gives: ac : Sc = sin a : sin 0, cb : Sc = sin 7 : sin 5. 2 This reasoning rests on the theorem, "If, from any point in the circumference, a perpendicular is dropped on a diameter of a circle, the perpendicular is the mean proportional of the segments deter- mined on the diameter," and its converse. 10 I. THE STEREOGRAPHIC PROJECTION Applied to the triangles ACD and BCD, the same law gives: sin a : sin ft = CD : CA, sin 7 : sin 5 = CD : CB. Since CA = CB (tangents drawn to the sphere from the same point), these ratios of sines are equal. Whence ac = c6, and c is the center of the projected circle. 3 FIG. 4. Center of projected circle. Note that if the given circle is a great circle (but not a meridian, nor the equator itself) its projection will be a circle having a radius larger than that of the equator and cutting the equator at the ends of a diameter. A great circle cuts the equator at the ends of a diameter of the latter; points on the equator are themselves their own stereographic projections. 3. Second Property of the Stereographic Projection. The angle between two circles is projected in true magnitude.* The 8 The proof holds good if the given circle is a great circle; the right cone with circular base becomes a right cylinder with circular base. Make the construction. 4 A more general property of the stereographic projection is that the angle between any two curves on the sphere is projected in true magnitude. The property proved in the text, however, is sufficient for our purpose. This proof can be omitted in a first reading. SECOND PROPERTY 11 angle between two circles drawn on a sphere is equal to the angle between their tangents. We shall prove: (1) that the angle between the projected tangents is equal to the angle between the tangents; (2) that the projected tangents are tangent to the projected circles. Thus will be established the property that the angle between the projected circles is equal to the angle between the circles. Fia. 5. Angle between projected tangents equal to angle between tangents. (1) Consider (Fig. 5) the point P in which two given circles intersect. Let PT and PT' be the tangents to these circles; they cut the plane of the equator in <, t' and the plane tangent to the sphere at S in T, T'. Join PS, intersecting the plane of the equator in p y the stereographic pole of P. The pro. jected tangents are pi, pt f . Join ST and ST'. The triangles TPT' and TST' are similar (TT common; TP = TS and TT = T'S, as tangents drawn to the sphere from the same point). Hence, angle TPT' = angle TST'. Again, the triangles tpt' and TST' are similar (all sides parallel each to each; two parallel planes being intersected by any 12 I. THE STEREOGRAPHIC PROJECTION third plane along parallel lines). Hence, angle tpt' = angle TST'. It follows that the angle between the tangents (TPT') and the angle between the projected tangents (tpt') are equal. s FIG. 6. Projected tangent, tangent to projected circle. *(2) Consider (Fig. 6) a right cone tangent to the sphere along the given circle PBC; let A be the vertex of this cone. We know that the circle PBC is projected as a circle pbc. We have seen that the projection of the vertex A is the center a of the projected circle. A generatrix AP of the right cone is projected as a radius ap of the projected circle. Pt, the tangent to the given circle at P, is projected in pt (t in the plane of the equator). But the angle apt, being the angle between the projections of the tangents PA and Pt, is equal to the angle between the tangents themselves, that is to say 90. Hence, pt is tangent to the projected circle. * Proofs believed to be new are marked by asterisks. CHAPTER II CESARO'S KEY-TRIANGLES 1 1. Ces&ro's "Triangle of Elements." Consider a spherical triangle ABC. Without loss of generality, we may suppose that one of its vertices A is located at the North pole of the sphere. Project this triangle stereographically. The pro- jected triangle A'B'C' (Fig. 7) will have a vertex A' at the FIG. 7. Stereographic projection. center of the projection, and two of its sides, A'B' and A'C', being projected meridians, will be straight lines; its third side B'C' will be an arc of a circle. At B' and C' draw the tangents to the circle B'C', meeting in T. Because the stereographic projection is angle-true, C'A'B' = A, A'B'T = B, A'C'T = C. 1 Ces&ro, G. Nouvelle mthode pour l^tablissement des formules de la trigonometric sphe*rique. Bull. Acad. roy. de Belgique (Cl. des Sc.), 1905, 434. Les formules de la trigonometric sphe"rique de"duites de la projection stereographique du triangle. Emploi de cette projection dans les recherches sur la sphere. Bull. Acad. roy. de Belgique (Cl. des Sc.), 1905, 560. 13 14 II. CESARO'S KEY-TRIANGLES Designate by 2E the external angle between the tangents meeting in T. It is easy to see that A + B + C - 180 2E. The angle 2E is called the spherical excess of the spherical triangle ABC. It is equal to the excess over 180 of the sum of the angles of the spherical triangle. We shall express it in degrees. The angles of the plane triangle A'B'C' are expressed as follows, in terms of the angles of the spherical triangle ABC and its spherical excess 2E, A' = A, B' = B - E, C' = C - E. The sides of A'B'C' are functions of the sides of the spher- ical triangle ABC (Fig. 8). Taking the radius of the sphere FIG. 8. Perspective drawing of the sphere of projection . as the unit of length, we have c' = tan - , b' = tan - . "TRIANGLE OF ELEMENTS" 15 Each of the quadrilaterals ABB'A' and ACC'A' has two opposite angles equal to 90 (one at A', by construction; the opposite one, as being inscribed in a semi-circle) and is, therefore, inscribable in a circle. Hence SB. SB' = SA.SA' = SC.SC' and BCC'B' are also concyclic. It follows that the angles .In I Fio. 9. Triangle of elements relative to the angle A. marked /3 (and the angles marked 7) are equal, so that the triangles SBC and SC'B' are similar. We may write, therefore, B'C' : BC = SC' : SB, or 2 b , sec H a 2 2 sin jr 2 cos whence a= c cos cos 8 For those who prefer step-by-step derivations: B'C' a', by definition. BC = 2 sin Ja, for the chord BC subtends an arc a and the chord is equal to twice the sine of half the angle. SC', in the right-angled triangle SC'A', where SA' 1, is the secant of the angle A ? SC'. Finally, in the triangle SAB, where the angle at B is a right angle, SB - SA cos ASB - 2 cos Jc. 16 II. CESARO'S KEY-TRIANGLES The triangle obtained by multiplying the three sides of A'B'C' by b c cos cos ^ is Cesaro's triangle of elements relative to the angle A (Fig. 9). Other "triangles of elements/* relative to the angles B and C, can be obtained by cyclic permutations. 2. Ces&ro's "Derived Triangle." A lune is the spherical sur- face bounded by two great semi-circles; for instance (Fig. 8), FIG. 10. The two complementary trihedra, each showing its six parts and spherical excess. the area ABSCA between two meridians, ABS and ACS. Two trihedra are called complementary when the two spherical triangles they determine on the sphere form a lune. For in- stance (Fig. 8), the trihedra A'ABC and A'SBC are comple- mentary. They have two edges, A'B and A'C, in common and the third edge, A'S, of one is the prolongation of the third edge, A A', of the other; the vertices A and S lie at the ends of a diameter, and the two spherical triangles ABC and SBC are seen to form a lune. Designating by O the center of the sphere, consider a trihedron OABC, represented (Fig. 10) by its spherical tri- angle ABC. Produce the great circles BA and BC till they meet, in D, thus forming a lune. The spherical triangle ADC represents the complementary trihedron OADC. The parts of 'DERIVED TRIANGLE* 1 17 the triangle ADC are easily expressed in terms of those of the triangle ABC. One side, 6, is common; the angle at D is equal to B; the other parts are the supplements of corre- sponding parts of the triangle ABC (Fig. 10). The spherical excess is found to be (180 - A) + (180 - C) + B - 180 = 180 - (A + B + C) + 2B = 2B - 2E = 2(B - E). cos 2 cos I FIG. 11. Derived triangle relative to the angle A. Let us compose the triangle of elements, relative to the angle (180 A), of the complementary trihedron OADC. Its six parts are tabulated below, together with those of the triangle of elements, relative to A, of the primitive trihedron OABC. THE SIX PARTS OF THE TRIANGLE OF ELEMENTS For trihedron OABC For trihedron OADC Angles Sides Angles Sides A BTT . a /1OAO A\ . 180 - a c . b c V.loU A) Bfft TT<\ . b 180 - CT7 1 sin cos 7: . c b /1ono /~i\ /"R Tp\ . 180 - c b 2 sin cos ~ 18 II. CESARO'S KEY-TRIANGLES The expression of the third angle is transformed as follows: 180 -C-B+E=A+ 180 -(A + B + C) + E = A 2E + E = A E. The other parts are easily simplified. The new key-triangle can now be drawn (Fig. 11); it is Cesar o's derived triangle relative to the angle A. The derived triangle of a trihedron is the triangle of elements of the complementary trihedron. Other "derived triangles/' relative to the angles B and C, can be obtained by cyclic permutations. 3. Polar Triangles. Consider a spherical triangle ABC and the corresponding trihedron OABC. Erect OA*, OB*, OC*, perpendicular to the faces of the trihedron, OBC, OCA, OAB, and on the same sides of these faces as OA, OB, OC, respec- tively. The new trihedron OA*B*C* determines, on the sphere (Fig. 12), a triangle A*B*C*. The vertices, A*, B*, C*, are called the poles of the planes OBC, OCA, OAB, respectively. C* FIG. 12. Polar triangles. The triangle A*B*C* is said to be the polar triangle of ABC. It follows from the construction that ABC is the polar triangle of A*B*C*. Likewise, it follows that the angles of one tri- angle are the supplements of the sides of the other, and that the sides of one are the supplements of the angles of the other. POLAR TRIANGLES 19 Hence the perimeter 2p* of A*B*C* is equal to 360 - 2E, and its spherical excess 2E* is equal to 360 2p, where 2E and 2p refer to the triangle ABC. In other words half the spherical excess of one triangle is the supplement of half the perimeter of the other. Both the triangle of elements and the derived triangle of the polar triangle A*B*C* are thus easily established (Figs. 13 FIG. 13. Triangle of elements of the polar triangle (relative to a). FIG. 14. Derived triangle of the polar triangle (relative to a). and 14). Their sides and angles are functions of the parts of the triangle ABC, in particular of its semi-perimeter p. Analogous key-triangles, relative to b and c, can be obtained by cyclic permutations. Remark. It is easy to remember how to transform the key-triangles of the primitive triangle into those of the polar triangle. Instead of a function of a half-side, write the 20 11. CESARO'S KEY-TRIANGLES a A co-function of the half-angle; thus sin ^ becomes cos ^ , Z 1 sin x cos jr becomes cos ^ sin -^ , etc. Instead of an angle, z & & write the supplement of the side; thus A becomes (180 a), and (180 A) becomes a. Instead of half the spherical ex- cess, E, write the supplement of half the perimeter (180 p). Instead of an angle minus half the spherical excess, write half the perimeter minus the side; thus (A E) becomes (p a) etc. The latter transformation is immediately apparent, since A* - E* = (180 - a) - (180 - p). 4. Derivation of the Formulae of Spherical Trigonometry. All the formulae of spherical trigonometry are derived from the key-triangles (which have been obtained without any restrictive hypothesis on the spherical triangle, and are there- fore perfectly general) by applying to them the known formu- lae of plane trigonometry. Each formula of spherical trigo- nometry can thus be derived independently of the others. FIG. 15. A plane triangle. A' + B' + C' = 180 a' + b' + c' = 2p' The parts of a plane triangle (Fig. 15) will be designated by primed letters: A', B', C', the angles; a', 6', c', the opposite sides f 2p' = a' + V + c', the perimeter. The parts of a spherical triangle (Fig. 16) will be desig- nated by unprimed letters: A, B, C, the angles; a, 6, c, the 21 FIG. 16. A spherical triangle. A + B + C - 180 = 2E a + b + c = 2p sides; 2p = a + b + c, the perimeter; 2E = A + B + C 180, the spherical excess. EXERCISES 1. Derive the triangle of elements relative to the angle B from that relative to the angle A (Fig. 9) by cyclic permutations. 2. Same question for the triangle of elements relative to C. 3. Draw the derived triangles relative to the angles B and C. 4. Derive, for the polar triangle, the triangle of elements and the derived triangle: (i) relative to the side b, (ii) relative to the side c. 5. Show that the area of the triangle of elements and that of the derived triangle are both equal to A/8, where A =* 2Vsin p sin(p a) sin(p b) sin(p c). (A, called the sine of the trihedral angle, is a useful function of the face angles of the trihedron, which are the sides of the spherical triangle.) CHAPTER III HOW THE KEY-TRIANGLES ARE PUT TO WORK 1. Napier's Analogies. 1 Napier's analogies are relations be- tween five parts of a spherical triangle. tan*(B-C) sin (& c) cot JA sin J(6+c)' tan %(b c) sin i(B-C) tan ia " sin J(B + C)' tan J(B-f C) cos i^c). cot JA cos Kb+c)' tan l(b-f-c) cos l(B-C) tan Ja cos i(B+Q* (1) (2) These formulae are read directly from the key-triangles, by applying the Law of Tangents: In a plane triangle, the tangent of the half-difference of two angles is to the tangent of their half-sum (or to the cotangent of half the third angle) as the difference of the opposite sides is to their sum. The triangle of elements (Fig. 9) yields the first analogy. Note that the half-sum of two angles is B C. We have . tan B C . b c . c b . b - c sin ^ cos ^ sin - cos ~ sm -^ .A cot .6 c , . c b sin ^ cos ^ -f sm ^ cos ~ sm b + c' The derived triangle (Fig. 11) gives the second analogy. The last two are obtained from the key-triangles (Figs. 1$ and 14) of the polar triangle. 2. Delambre's Formulae. 2 Delambre's formulae are rela- tions between all six parts of a spherical triangle. 1 Analogies is an archaic term for proportions. 1 Improperly attributed to Gauss by certain authors. NAPIER'S AND DELAMBRE'S FORMULAE 23 co8j(B-C) ^ sin j(b+c) sin JA sin Ja ' cos J (B + C) ^ cos j(b-f c) sin JA cos Ja ' sin i(B-C) sin J(b c) > cos JA sin Ja ' sin i(B+C) cos J(6 c) cos JA cos Ja (3) (4) The Law of Sines of plane trigonometry, applied to the triangle of elements (Fig. 9), gives sm - . 6 c sin -cos . c b am cos sin A sin (B - E) sin (C - E) ' which, by the theory of proportions, becomes . b + c Sm :r sm b - c . a sm o 2 _ z z sin A ~ sin(B-E) + sin(C-E) ~ sin(B-E) - sin(C-E) or a .b + c .... 6 c sm 2 sm sm .A A A B-C 2 Sm TT COS TT 2 COS 77 COS S . A . B - C' 2 sm " sm and Delambre's first two formulae follow immediately. The same method, applied to the derived triangle (Fig. 11), yields the last two formulae. 3. Euler's Formula. Euler's formula expresses E, one-half of the spherical excess, 3 in terms of the sides. cos E = cos a -f cos b + cos c . a b c 4 cos cos ~ cos ~ (5a) 3 The value E of one-half the spherical excess is useful in calculating the area of a triangle. It is known from geometry (see Appendix 1) that the area of a spherical triangle is to the area of the sphere as E (expressed in degrees) is to 360. 24 III. KEY-TRIANGLES PUT TO WORK This formula is obtained from the derived triangle (Fig. 11), by applying the Law of Cosines of plane trigonometry a '2 = /2 + c > 2 _ ^V c' cos A'. We have . 6 . n c n d , 6 c rt a b c r. sm 2 sin 2 - = cos 2 ~ + cos 2 ~ cos 2 ~ 2 cos cos ~ cos ~ cos E, Z Z t & & & & & n a b c ^ 9 a 2 cos cos ~ cos ~ cos E = cos 2 ~ JL u ,/ 6 c . . 6 . c \/ 6 c . b . c\ + ( cos ^ cos g + sm - sm - 1 ( cos ^ cos ^ - sm 3 sm 2 / ' A a b c -n o 9 a .o & C 6 + C 4 cos H cos jc cos cos E = 2 cos 2 + 2 cos cos ^ , & & A & A from which Euler's formula follows immediately. 4. Lhuilier's Formula. Lhuilier's formula gives the fourth of the spherical excess in terms of the sides (and the semi- perimeter p). x oE , p . p a , p b , p c tan 2 ^ = tan ^ tan ^r tan ^ ^ tan *-= L t & fj t (5b) It can be obtained directly from the derived triangle (Fig. 11) by expressing the tangent of half the angle E in terms of the sides. The plane trigonometry formula is - p'(p' - a') We have o f OL/ 6 c a 2p 26 = cos ~ -- cos ~ , /u t r / 6 c , a 2p' = cos -- + cos , FORMULAE FOR SPHERICAL EXCESS 25 and r I r / a b + C 2p 2c = cos K cos ^ . Since 2p' 2a' = cos ^ + cos ^ , cos P - cos Q , P + Q , Q - P - - - = tan 1 r^- tan -* - - - 1 r- -= , cos P + cos Q 2 2 * we are led to , E . a + b c L a b + c tan 2 75- = tan - - - tan - - - . . . a + b 4- c , b + c a X tan - -r - tan - - - , which is Lhuilier's formula, as 2p = a + b + c. *5. Cagnoli's Formula. Cagnoli's formula gives the half of the spherical excess in terms of the sides (and the semi- perimeter p). . ^ Vsm p sm(p a) sin(p 6) sm(p c) sin Hi = r a b c 2 COS jr COS COS 7 (5c) The area T' of the derived triangle (Fig. 11) may be expressed as the half-product of the base by the altitude, T' = ? cos 5 cos ~ cos ~ sin E, u i t or, in terms of its sides and semi-perimeter, 4 = i Vsin p sin(p a) sin(p b) sin(p c). 4 Cp. Exercise 5, Chapter II and Section 4, this chapter. 26 III. KEY-TRIANGLES PUT TO WORK Equating the two expressions immediately gives the desired formula. *6. The Spherical Excess in Terms of Two Sides and Their Included Angle. In a plane triangle, as a consequence of the Law of Sines, c' - V cos A' cot B' = b' sin A' This formula, applied to the derived triangle (Fig. 11), in which the angles (180 A), E, (A E) are taken as A', B', C', respectively, yields the desired relation , T-, cot E b c , . 6 . c . cos x cos JJT + sin ~ sin ^ cos A . 6 . c . A sin - sin jr sin A fL or cot - cot ~ + cos A cot E = 4^ sin A (6) EXERCISES 1. Express cosp in terms of the angles, by applying Euler's formula to the polar triangle. 2. Derive the expression of cos p in terms of the angles from the derived triangle of the polar triangle (Fig. 14), in the same way as Euler's formula has been derived (Sn. 3). 3. From Napier's second analogy (1), derive a formula to calculate E in terms of two sides (6, c) and their included angle (A) . 4. Show that, in a right-angled triangle (C = 90), tan E = tan - tan ^ . Hint: use the formula derived in the preceding exercise. E E 5. Find sin ^ and cos jr- in terms of the sides. Apply the s method as that used in Sn. 4 for deriving Lhuilier's formula (6). EXERCISES 27 6. Derive Lhuilier's formula (6) from the results obtained in the preceding exercise. 7. Derive Cagnoli's formula from Lhuilier's formula. 8. Gua's formula gives cot E in terms of the sides. Find what it is. 9. You now know six formulae by means of which, given the sides, the spherical excess can be calculated. Which one would you prefer if you had to depend on logarithmic calculations? Which one would be easiest to use if a calculating machine were available? 10. The sides of a spherical triangle are a = 3526', 6 = 4215', c 1822'. Calculate the spherical excess by two different formulae. CHAPTER IV RELATIONS BETWEEN FOUR PARTS OF A SPHERICAL TRIANGLE 1. Relation between the Three Sides and One Angle. 1 Ap- plying the Law of Cosines of plane trigonometry a'* = fc'2 + c '2 _ ^V c' cos A' to the triangle of elements (Fig. 9), we get a b c c b sin 2 = sin 2 ^ cos 2 ~ + sin 2 ~ cos 2 ~ . b c . c b . 2 sm cos H sin cos ~ cos A. In view of 2 sin 2 - = 1 cos x, 2 cos 2 ~ = 1 + cos x\ t A the above formula, multiplied by 4, may be written 2(1 cos a) = (1 cos b)(l + cos c) + (1 cos c)(l + cos b) 2 sin b sin c cos A, whence the desired formula ' (7) cos a cos b cos c + sin 6 sin c cos A. This is expressed: The cosine of the side opposite the given angle is equal to the product of the cosines of the other two sides, plus the product of the sines of these two sides times the cosine of the given angle. i Used in Ex. 16, 21, 24, 26, 28 (Ch. VIII). OBLIQUE-ANGLED TRIANGLES 29 2. Expression of the Half-Angles in Terms of the Three Sides. 2 The following formulae of plane trigonometry l 2 ' - V)(p' - c') 6V COS' A' _ p '(p> - fl ') 2 6V p'(p> - a') are applied to the triangle of elements (Fig. 9). We have n , . b + c . . a n . p p a 2p' = sm ^ -- h sin - = 2 sm | cos ^ , 2p' 2a' = sin .a n . p a p , sm ^ = 2 sin - - cos ^ , etc. Hence, in the spherical triangle, . 2 A sm ^ 2 A cos 2 2 - tan*^ sin(p - 6) sin(p - c) (8a) (8b) (8c) sin 6 sin c ' sin p sin(p a) sin 6 sin c ' 2 sin p sin(p a) These formulae are easily remembered on account of their similarity with the corresponding formulae of plane trigo- nometry. 3. Relation between Two Sides and Their Opposite Angles. 3 Apply the Law of Sines of plane trigonometry to the derived triangle relative to A (Fig. 11). a . b . c cos sm sm sin A sinE Used in Ex. 1, 18, 19, 20 (Ch. VIII). 3 Used in Ex. 18, 21, 25, 27 (Ch. VIII). 30 IV. FOUR PARTS OF A SPHERICAL TRIANGLE Multiply both members by 2 sin - , i . a . b . c 2 sin pr sm - sin s sin a _ 222 sin A sin E In like manner, from the derived triangle relative to B, we get n . a . b . c . , 2 sin jr sin ^ sin ^ sin 6 __ 222 sin B sin E * which was to be expected from the symmetrical form of the right-hand member of the above equations. Hence sm a __ sin b sin A ~~ sin B ' (9) This is expressed: The sines of the sides are proportional to the sines of the opposite angles. 4. Relation between the Three Angles and One Side. 4 The Law of Cosines of plane trigonometry is applied to the triangle of elements of the polar triangle (Fig. 13). The method is the same as for the first formula (Sn. 1). cos A = cos B cos C + sin B sin C cos a. (10) This formula is easily remembered on account of its simi- larity with formula (7). Note the minus sign in the second member, however. 5. Expression of the Half-Sides in Terms of the Three Angles. 6 The method used in Sn. 2 could be applied to the triangle of elements of the polar triangle (Fig. 13). Another Used in Ex. 27 (Ch. VIII). Used in Ex. 19 (Ch. VIII). OBLIQUE-ANGLED TRIANGLES 31 method does not require the use of the polar triangle. Draw the two key-triangles relative to B and those relative to C. The Law of Sines of plane trigonometry, applied to the triangle of elements relative to C, gives sin(A - E) sin C . a b sm cos sn the same law, applied to the derived triangle relative to B, gives sin E sin B . c . a Sin 2 sm 2 b~ COS?: Multiplying these two relations by each other yields immedi- ately , a sin E sin(A E) i .. 2 sin B sin C (lla) Likewise, the Law of Sines, applied to the triangles of elements relative to B and to C, yield two relations: one between B, C E, and the opposite sides; the other, between C, B E, and the opposite sides. These two relations, multiplied by each other, give cos sin(B - E) sin(C - E) sin B sin C (lib) Finally, the Law of Sines may be applied to both key- triangles relative to B, giving the relations sin E . c . a sm sm sin(B - E) c a ' cos cos ~ . a c / A T-\ sin r cos o sm(A - E) 2 2 sin(C - E) ~ . c a ' v 7 sm H cos H 32 IV. FOUR PARTS OF A SPHERICAL TRIANGLE which, multiplied by each other, lead to 2 a __ sin E sin(A E) 2 ~ sin(B - E) sin(C - E) ' (He) The last formula could, of course, be derived from the pre- ceding two. The formulae (11) can only be remembered after careful comparison with the formulae (8). Note the deceiving simi- larity between (8a) and (lib), etc. 6. Relation between Two Sides, their Included Angle, and the Angle Opposite One of them (that is to say, between four consecutive parts). 6 Napier's first two analogies, as read from the key-triangles relative to C, give . a - b f A-B Sin ~2~ .0 tan ___ = __ n - cot _, sin tan A + B cos- 2 a b C cos A relation between a, 6, C, A can be obtained from these two equations by eliminating B, which is easily done as follows. Since A-B , A+B A - A. we may write tan A tan 2 A - B + tan . 1 tan A - ^ tan 2 ' A + B 2 _ N A + B ~ D' Used in Ex. 2, 21, 27 (Ch. VIII). OBLIQUE-ANGLED TRIANGLES 33 where r . g-6 sin _ N = cos a - b} sm cos a + b cot - o 2 sm a cos ~- L _ sin(a + b) sin ^ sin a sin C Q sin (a + 6) sin 2 =- and sin(a - b) C D = 1 -- rf - r-TT COt 2 -pr sm(a + 6) 2 C C sin(a + b) sin 2 -^ sin(a 6) cos 2 7^- 2i 2t = _ sin (a + 6) sin 2 TT- J Substituting for N and D gives tan A __ sin a sin C a cos 6 ( sin 2 ~ cos2 o") +sin6cosaf sin 2 ^ sn whence sin 6 cos a sin a cos 6 cos C = sin a sin C cot A, or, on dividing by sin a and transposing, cot a sin b = cos 6 cos C + sin C cot A. (12) 34 IV. FOUR PARTS OF A SPHERICAL TRIANGLE This is known as the "cot-sin-cos" formula. (Note the symmetry in the sequence of the trigonometric functions: cot-sin-cos . . . cos-sin-cot.) It is expressed: The cotangent of the side opposite one of the given angles times the sine of the other side is equal to the cosine of the latter times the cosine of the included angle, plus the sine of the latter times the cotangent of the angle opposite the first side. Remark. Formula (12) is sometimes written in another, just as elegant, form: cos b cos C = sin b cot a sin C cot A. (12a) It is then expressed: The product of the cosines of the side and the angle that are not opposite any given part is equal to the difference of their sines, each multiplied by a cotangent, respectively that of the given side and the given angle that are opposite each other. EXERCISES 1. Express cosp in terms of the angles, by using Delambre's formulae and formula (7). Hint: P = Ja 4- J(& + c). 2. Transform formula (lib) into formula (10). Hint: To elimi- nate E multiply both members by 2 and change the product of two sines in the numerator into a difference of two cosines. 3. Derive formula (10) by applying formula (7) to the polar triangle. 4. Derive the three formulae (11) by means of the triangle of elements of the polar triangle. 5. Derive formula (12) from Napier's last two analogies (2). 6. Write the formula giving cos c in terms of o, 6, C. Then replace cose in tne formula (7) by the value just found and, using the proportionality of the sines of sides and opposite angles, derive formula (12). 7. Let 2S designate the sum of the angles of a spherical triangle. What form will the three formulae (11) assume with this notation? EXERCISES 35 8. From formulae (8a) and (8b), find an expression of sin A. Show that it is equivalent to 1 sin A : Vl cos 2 a cos 2 6 cos 2 c + 2 cos a cos 6 cos c. sin 6 sin c *9. The expression of sin A, in terms of the sides, found from formulae (8a) and (8b) can also be obtained directly from the triangle of elements (Fig. 9), by applying to it the method used for deriving Cagnoli's formula (Ch. Ill, Sn. 5). CHAPTER V RIGHT-ANGLED TRIANGLES 1. The Right-Angled Triangle, a Special Case of the Oblique- Angled Triangle. The formulae established in the preceding chapter give relations between four parts of a triangle. Of these parts, at least one is necessarily an angle. By letting an angle equal 90, a formula derived for the general case of an oblique-angled triangle is transformed into a relation be- tween three parts (other than the right angle) of a right-angled triangle. By convention l the triangle ABC is made right- angled at C. It may be necessary to rearrange the general formulae by permutation of letters, so that the angle which is to become the right angle be labeled C. The side opposite C will therefore be the hypotenuse c. There are six different ways of choosing three parts from the five parts (other than the right angle) of a right-angled triangle. The hypotenuse may be taken, either with the other two sides, or with the two angles other than the right angle, or with one side and one angle. In the latter case, the angle may be opposite the chosen side or adjacent to it. If the hypotenuse is not chosen, there are only two possibilities : two angles and one side, necessarily opposite one of the angles; or two sides and one angle, necessarily opposite one of the sides. Right-angled triangles are always with us. The six rela- tions about to be derived have therefore proved to be very useful in practice. You will find it advantageous to commit them to memory, preferably in the form of statements rather than equations. 1 This convention is not universal. Many authors make A = 90. 36 PRODUCTS AND RATIOS 37 2. Relations between the Hypotenuse and Two Sides or Two Angles. 2 The Product Formulae. Formula (7) may be written cos c = cos a cos 6 -f sin a sin b cos C. It becomes, 8 for C = 90, cos c = cos a cos b. (13) Formula (10) may be written cos C = cos A cos B + sin A sin B cos c. It becomes 4 cos c = cot A cot B. (14) This is expressed: The cosine of the hypotenuse is equal to the product of the cosines of the two sides, or the product of the cotangents of the two angles. 3. Relations between the Hypotenuse, One Side, and One Angle. The Ratios of Sines and Tangents. Formula (9) may be written sin c sin a sin C sin A " It becomes, 5 for C = 90, sin a = sin c sin A J The terms sides and angles, when applied to a right-angled tri- angle, are construed to mean sides other than the hypotenuse and angles other than the right angle. Formula (13) is used in Ex. 3, 7, 15, 17, 29 (Ch. VIII). * Formula (14) is used in Ex. 9, 12, 26 (Ch. VIII). 5 Formula (15) is used in Ex. 4, 14, 23, 24, 29 (Ch. VIII). 38 V. RIGHT-ANGLED TRIANGLES or sin a sin c = sin A. (15) Formula (12) may be written cot c sin a = cos a cos B + sin B cot C. It becomes 6 cos B = cot c tan a or (16) With respect to the side a, note that A is the opposite angle, whereas B is the adjacent angle. The relations are remembered as follows. In each case consider the side and the hypotenuse; the ratio of their sines is equal to the sine of the opposite angle, the ratio of their tangents is equal to the cosine of the adjacent angle. Remark. Compare the definitions of sine and cosine in a plane right-angled triangle: sin A' = cos B' = a'/c'. 4. Relations between Two Angles and One Side, or be- tween Two Sides and One Angle. The Ratios of Cosines and Cotangents. Formula (10), cos A = cos B cos C + sin B sin C cos a, becomes, 7 on letting C = 90, cos A = sin B cos a or (17) Formula (16) is used in Ex. 4, 8, 20 (Ch. VIII). 7 Formula (17) is used in Ex. 13 (Ch. VIII). NAPIER'S RULE Formula (12), cot a sin b = cos b cos C + sin C cot A, becomes, 8 for C = 90, cot a sin b = cot A or (18) The relations are remembered as follows. In each case consider an angle and its opposite side; the ratio of their cosines is equal to the sine of the other angle, the ratio of their cotangents is equal to the sine of the other side. Remark. Notice that all but one of the four ratios con- sidered in the last two sections give the sine of a part. 5. Napier's Rule. Although some people prefer to memorize the preceding formulae as such, Napier's rule, which includes all the possible relations between any three parts of a right- angled triangle, may be found useful by others. FIG. 17. Napier's rule. In the triangle ABC, right-angled at C, ignore the right angle; then, the hypotenuse, the other two angles, and the 8 Formula (18) is used in Ex. 7, 15, 17 (Ch. VIII). 40 V. RIGHT-ANGLED TRIANGLES complements of the sides about the right angle are Napier's five cyclic parts. The rule is expressed as follows. The cosine of any of the five parts is equal to the product of the sines of the opposite parts or the product of the cotangents of the adjacent parts. cos c = cos a cos b = cot A cot B, cos A = sin B cos a = cot c tan b, cos B = sin A cos b = cot c tan a, sin a = sin c sin A = cot B tan &, sin 6 = sin c sin B = cot A tan a. This amazing rule is more than just a mnemonic trick. It is a theorem, which was given with a separate proof 9 by Napier. It is easy to check that the ten relations yielded by the rule correspond, with some duplication, to the formulae (13) to (18). * Napier's proof is outside the scope of this book. CHAPTER VI EXAMPLES OF CALCULATIONS 1. Solution of Triangles. 1 (i) Oblique-angled triangles. Three parts being given, any fourth part may be calculated by one of the formulae (7) to (12). To find an angle in terms of the three sides, use one of the formulae (8), giving the half- angle. Likewise, to find a side in terms of the three angles, use one of the formulae (11), giving the half-side. In all other cases, use one of the fundamental formulae (7), (9), (10), (12). (ii) Right-angled triangles. Two parts being given, besides the right angle, any third part may be calculated by one of the formulae (13) to (18). Find the appropriate formula or use Napier's rule. An isosceles triangle is divided into two right-angled tri- angles by an arc drawn from the vertex at right angles to the base; this arc bisects the base and the opposite angle. Isos- celes triangles may also be solved like oblique-angled triangles by means of the fundamental formulae; formulae (8) and (11) should be avoided, as they would complicate the calculations. (iii) Right-sided (or quadrantal) triangles. The polar triangle of a right-sided triangle is right-angled; Napier's rule may be used to solve the polar triangle, from which the parts of the quadrantal triangle are then computed. 1 We have seen (Introduction, Sn. 2) that any angle or any side of a spherical triangle is less than 180. It follows that a case of am- biguity will present itself whenever a part is to be calculated by a formula that gives its sine (two supplementary angles having equal sines). 41 42 VI. EXAMPLES OF CALCULATIONS Right-sided triangles are also easily solved by the funda- mental formulae, which are greatly simplified when one of the sides is equal to 90. 2. Formulae Adapted to Logarithmic Computation. (i) FORMULA: cos a = cos b cos c + sin b sin c cos A. The unknown may be either a, b (or c), or A. If the un- known is a, factorize cos 6: cos a = cos 6(cos c -f tan 6 sin c cos A). Introduce an auxiliary angle u so as to arrive at the sine of the sum of two angles. Let cot u = tan b cos A, then cos b , . . . . cos a = (sm u cos c + sm c cos u) or cos 6 sin(c + u) cos a = r-^ ! - . sm u If the unknown is c, the same method gives immediately cos a sin u sin(c -f w) = cos b If the unknown is A, use the formulae (8), which are adapted to logarithmic computation. (ii) FORMULA: cot a sin 6 = cos b cos C -f- sin C cot A. If the unknown is a, we write A j. r / ~ , sin C cot A \ cot a = cot 61 cos C H r ) . \ cos 6 / Let . cot A cot v = r , cos 6 LOGARITHMIC COMPUTATION 43 then cot b , . ~ . . ~ x cot a = . - (sin v cos C + sin C cos v) sin t; or cot 6 sin(C + t;) - ~ - - . cot a = sin v If the unknown is C, the same method gives immediately cot a sin t; If the unknown is 6, we write A / r ,cosC\ . ~ , A cot al sin o cos b - 1 = sin C cot A. \ cot a / Let cos C tan w = 7 , cot a then COt d . 1 f \ SI ! A (sin o cos w sin w cos 6) = sm G cot A cos w or . ,, x sin C cot A cos w sm(6 w) = T . cot a If the unknown is A, the same method gives immediately , . cot a sin(6 w) cot A = : rr - . sin C cos w (iii) FORMULA: cos A = cos B cos C + sin B sin C cos a. The method is the same as for the first formula (Sn. 2, i). The results follow. Let cot x = tan B cos a. If the unknown is A, . cos B sin(C x) cos A = r-^ . sm x 44 VI. EXAMPLES OF CALCULATIONS If the unknown is C, . ,~ x cos A sin x sm(C x) = 5 . cos B If the unknown is a, use the formulae (11), which are adapted to logarithmic computation. 3. Numerical Calculations. Computations are carried out either by means of logarithms or with a calculating machine (in which case tables of natural values of the trigonometric functions are necessary). In either case, a neat calculation form is essential. (i) By logarithms. First write across the top of the page the formula to be employed (adapted to logarithmic computation, if desired). In this formula always designate the parts of the triangle by letters (not by their actual values). Then write the given parts near the left margin of the sheet. Avoid writing the word log as much as possible, use it only in front of those logarithms which may be needed again in the course of the calculations. Write the minus sign over a negative characteristic, 2 so as to avoid the cum- 1 Calculations with negative figures are quite straightforward, once you become used to them. See for yourself: 1.67 2.67 2)3.34 2)3.34 2__ _ 13 13 12 12 14 14 14 14 In the division on the left, you say: 2 in 3 goes 1, 1X2 = 2, 3 2 = 1, bring down 3, 2 in 13 etc. In the division on the right, you say: 2 in 3 goes 2, 2X2 = 1, 3 1 = 1, bring down 3, 2 in 13 etc. The only difference is that, in the second case, you take the first partial quotient by excess in order to get a positive remainder. FIRST EXAMPLE 45 bersome addition and subtraction of 10. In most cases there is no advantage in adding the cologarithm rather than sub- tracting the logarithm itself; in a simple division, especially, this amounts to making a subtraction and an addition instead of one subtraction only. Write the result near the left margin of the page. A second formula may be used as a check. This can be done in two ways: either the solution is carried out in dupli- cate, by means of two formulae; or a relation between the given parts and the result is verified afterwards. In the course of long calculations (involving a whole chain of tri- angles, for instance) it is well to check the results from time to time, before proceeding. (ii) With the calculating machine. Write the formula across the page. Copy only those natural values that may be used again in the course of the calculations. Write the data and the result near the left margin of the page. 4. First Example. Given the three sides of a triangle, solve for one angle. Use a check formula. Again compare these two additions of logarithms : 9.412 5062 -10 1.412 5062 8.803 7253 -10 2.803 7253 18.216 2315 -20 2.216 2315 The last partial addition, in the example on the right, reads: 1 (carried over) and I make 0, and 2 make 2. Is this really difficult? 46 VI. EXAMPLES OF CALCULATIONS (A) SOLUTION BY LOGARITHMS . , A sin (p 6) sin (p c) , A sin p sin(p a) Sin 2 -pr : - : COS 3 7j- == -. j : . 2 sin o sin c 2 sin o sm c a = 5048'20" b - 11644'50" (6315'10") T.950 8518 c - 12911'40" (5048'20") 1.889 3049 1.840 1567 2p = 29644'50" 171 p - 14822'25 // (3137'35") T.719 6275 p - a - 9734' 5" (8225'55") 1. 996 1989 14 T. 715 8449 -T.840 1567 <- A T.875 6882 log cos = 1.937 8441 2 8462 21 12 1 8 9 - 2955'41.7" i Z p - 6 - 3137'35" .. T.719 6446 p~c - 1910'45" .. T.516 5358 303 T.236 2107 -T.840 1567 <- A T.396 0540 log sin ~ = 1.698 0270 2 0204 66 36 6 29 4 - 2955'41.8" 6 A - 5951'23.6" The answer to the closest second is 5951 / 24 // . Suggestion To find the supplement of an angle, write down the figures from left to right as you mentally increase the FIRST EXAMPLE 47 given angle to 17959'(50/10)". Example: given 14822'25". Write 31 while saying 179, 37' while saying 59, 3 while saying 50, and 5" while saying 10. (B) SOLUTION BY NATURAL VALUES AND CALCULATING MACHINE . cos a cos 6 cos c COS A = - : r : - , sin o sin c cos sin a= 5048'20" .......... 0.6319542 .......... b 11644'50" .......... - 0.450 0551 .......... 0.893 0008 c = 12911'40" .......... - 0.631 9542 .......... 0.775 0058 cos A = 0.502 1668 1817 149 125 7 A 5951'23.6". 23 3 i e i A * A sin(p 6) sin(p c) Check formula: tan 2 ^ = ~ - r ( 1 ~ ; . 2 sin psm(p a) 2p = 29644'50" p = 14822'25" (3137'35") .................... 0.524 3575 p - a = 9734' 5" (8225 / 55 // ) ... 0.991 2859, . . . 206 _ 32 0.524 3781 0.991 2891 p - b = 3137'35" .............................. 0.524 3781 p - c 1910'45" .............. 0.328 5003 _ 229 0.328 5232 tan 8 = 0.331 4101 tan = 0.575 6824 2 6708 116 64 6 = 29 55'41.8" 51 4 L A - 5951'23.6 /; . 48 VI. EXAMPLES OF CALCULATIONS 5. Second Example. Given the three angles of a triangle, solve for the three sides. (A) SOLUTION BY LOGARITHMS . . a sin E sin(A - E) . . b sin E sin(B - E) Sin 2 p: = - : ^5-^ ~ - - , Sin 2 - = - : ^ - - - '- 2 sm B sm C 2 sm C sin A . . c sinEsin(C - E) Sin 2 - = - : - T : ^5 - . 2 sin A sm B A = 6647' 0" ... T.963 3253 ... T. 963 3253 B = 4230'40" . . T . 829 7752 ... T . 829 7752 C = 9720'30". .T.996 4249 ... T. 996 4249 _ 20638'10" T.826 2001 T.959 7502 T.793 1005 2E - 2638'10" E = 1319' 5". .T.362 3558 . . . T.362 4003 . . . 1.362 4003 445 A - E = 5327'55"..1.904 9760 78 B - E - 2911'35" .............. T.688 1819 188 C - E = 84 1'25" . .......... ____ .......... . . T.997 6331 T.267 3841 T.050 6010 T.360 0334 -1.826 2001 -1.959 7502 -T.793 1005 ,.1.441 1840 h T.090 8508 ,1.566 9329 log sin. ^T.720 5920 .5 T.545 4254 . T.783 4664 2 5834 2 3938 2 4575 86 316 89 68 2 280 5 82 8 17 8 35 5 62 = 3142'12.5" a - 6324'25.0 /; b - 41 6'31.2" c - 74 48 ; 6.4" 2p - 179 19' 2.6" Check formula : tan 2 -^ = tan ^ tan tan tan ^r i & & & & SECOND EXAMPLE 49 p - 8939'31 . 3" ip - 4449'45 . 6" .... T . 997 3891 211 25 3 p - a - 2615' 6.3" J(p - a) - 13 7'33.1" T.367 6676 285 6 9 5 p - 6 - 4833 / 0.1" i(p - 6) - 2416'30.0" T.654 1690 p-c= 1451'24.9" |(p-c) 725'42.4" T. 115 1788 328 65 6 F 2.1344970 F log tan ^ = 1.067 2485 ~ = 639'32.4" 2 2041 444 E = 1319' 4.8" 366 78 (fi) SOLUTION BY NATURAL VALUES AND CALCULATING MACHINE cos A + cos B cos C , cos B + cos C cos A COS a = : ^ : 7= . COS = : -^T-- 1 > sin B sin C sin C sm A cos C -f cos A cos B COS C = : 1 : r; . sm A sin B cos sin A 6647' 0" 0.394 2093 0.919 0207 B = 4230'40" 0.737 1463 0.675 7332 C = 9720'30" -0.127 7859 0.991 8018 20638 / 10 // a = 6324'25.0" 0.447 6505 6724 219 217 2 6 - 41 6'31.3" 0.753 4637 4678 41 31 9 9 1 c 7448' 6.5" 0.262 1589 1892 303 280 8 22 2 50 VI. EXAMPLES OF CALCULATIONS Check formula: E as * -f cos a -f- cos 5 4- cos c A a b c 4 cos ~ cos o cos ~ 0.850 7602 177 8 12 7 Ja 3142'12.5" 0.850 7792 0.936 3322 68 6 8 i& 2033'15.6" 0.936 3397 0.794 3852 176 4 23 5 Jc 3724' 3.2" 0.794 4052 cosE = 0.973 1061 1119 58 56 E = 1319 ; 5.2" 2 2E 2638 / 10.4". 6. Third Example. In a triangle ABC, right-angled at C, the following parts are known: a = 5936 / 30 // , b = 6422'. Find the angle A. (A) SOLUTION BY LOGARITHMS sin b = cot A tan a, hence tana tan A sin 6 o - 5936 / 30" 0.231 7312 6 6422 ; -T.955 0047 log tan A - 0.276 7265 167 98 A - 62 7'52". GEOGRAPHICAL PROBLEM 51 (B) SOLUTION BY NATURAL VALUES . tana 1.705 0269 , QM 1COO tan A -T-T- = ^ nn* EQI^ "* 1 .891 1522 sin 6 0.901 5810 ~428 A = 62 7'52". 7. Fourth Example. Given the geographical coordinates of two points A and B on a sphere, find the spherical distance between these two points. Let L and M, L' and M' be the longitudes and latitudes of the points A and B, respectively (Fig. 18). Let x be the arc AB. FIG. 18. Geographical problem. The formula (7), applied to the triangle NAB, immediately solves the problem. cos x = sin M sin M' + cos M cos M' cos(L L'). If the calculation is to be made by logarithms, write cos x = sin M'[sin M + cos M cot M' cos(L L')]. Let cot M' cos(L L') = tan w and compute the auxiliary angle w. 52 VI. EXAMPLES OF CALCULATIONS We have then sin M' . _ .. , .,,. . s cos x = (sin M cos w + cos M sm w) cos w _ sin M' sin(M + w) cos w 8. Numerical Application. Take the following geographical coordinates: New York, lat. 4043'N., long. 740'W.; San Francisco, lat. 3748'N., long. 12224'W. Assume the earth to be spherical and take its radius to be 3960 miles. What is the spherical distance (in degrees and minutes) between New York and San Francisco? What is the actual distance (in miles)? The coordinates being accurate to half a minute, with what precision can you give the distance? (A) SOLUTION BY LOGARITHMS Let L, L' be the longitudes of San Francisco and New York. Let M, M' be their latitudes. j.** f /T T f \ sin M'sin(M -f- w) tan w = cot M' cos(L L ), cos x = . v ' cos w M' 4043' 0.065 1775 L 12224'30" L' 74 (T 0" L - L' 48 24'30" T.822 0487 T.887 2262 2024 238 217 5 20 5 w 3738'35.5" T.898 6243 M 3748' 65 2 M + w 7526 / 35.5 // 8 2 log cos w = 1.898 6316 NUMERICAL APPLICATION 53 M' T. 814 4600 M + ti; T.985 8270 27 5 2 8 T.800 2900 - 1.898 6316 log coax = T.901 6584 6649 65 x = 37 7'14.1" 636 1 4 The answer to the nearest half-minute is 37 7'. Length of the great circle on the earth (R = 3960 miles). 2 0.301 0300 TT 0.497 1499 R 3.597 6952 log27rR = 4.395 8751 8678 73 70 2wR = 24,881. 42 miles 3 Length of 1, 1', 1" _ _ __ 27rR 4.395 8751 4.395 8751 4.395 8751 360 -2.556 3025 21,600' -4.334 4538 1,296,000" -6.112 6050 1.8395726 0.0614213 2.2832701 37 1.568 2017 V 0.8450981 14.1" 1.149 2191 3.407 7743 0.906 5194 T.432 4892 7647 82 83 96 12 37 - 2,557 . 25 The error on J' is more than i mile. 7 8 . 0634 14. 1" - 0.2707 The answer to the nearest half-mile 2,565.58 is 2565* miles. 54 VI. EXAMPLES OF CALCULATIONS (B) SOLUTION BY NATURAL VALUES AND CALCULATING MACHINE cos x = sin M sin M' + cos M cos M' cos(L L')- M 3748'.. 0.6129071 0.7901550 L = 12224'30" M' 4043' . . X 0.652 3189 X 0.757 9446 L' = 74 0' 0" X 0.663 8174 L - I/ = 4824'30" cosx = 0.797 3669 3792 123 117 2 3-37 7'14.2" 58 Tables give the lengths of arcs to the radius 1. 37 .... 0.645 7718 7 ; .... 0.002 0362 J' .... 0.000 1454 14 ;/ .... 0.000 0727 X 3960 0.6478807 J' = 0.5758 miles X 3960 x = 2565.6 The answer to the nearest half-mile is 2565 i miles. CHAPTER VII PROBLEMS 1. Let ABC be a tri-rectangular triangle (all angles and sides equal to 90) ; let p, q, r be the spherical distances from any point P inside the triangle to the vertices A, B, C, respectively (Fig. 19). Prove that cos 2 p + cos 2 q + cos 2 r = 1. 90, FIG. 19. Distances from a point inside a tri-rectangular triangle to the vertices. In the triangle CAP, by formula (7), cos p = sin r cos x. Likewise, in the triangle BCP, cos q = sin r sin x. Squaring and adding, in order to eliminate x, we get cos 2 p + cos 2 q = sin 2 r = 1 cos 2 r, whence the desired relation. 55 56 VII. PROBLEMS 2. A point P is located inside a tri-rectangular triangle ABC. Let x, y, z be its distances from the sides (Fig. 20). Prove that sin 2 x + sin 2 y + sin 2 2 = 1. 90 FIG. 20. Distances from a point inside a tri-rectangular triangle to the sides. Join PA, PB, PC and let these arcs be p, q, r. Let the arcs x, y, z intersect the sides BC, CA, AB in D, E, F. Let BD = u, CE = v, AF = w. The two right-angled triangles BDP and GDP yield the relations: cos q = cos x cos u t cos r cos x sin u. On squaring and adding, we get cos 2 x = cos 2 q + cos 2 r. Likewise, cos 2 y = cos 2 r + cos 2 p, cos 2 z = cos 2 p + cos 2 q. On addition, and in view of the fact that cos 2 p + cos 2 q + cos 2 r = 1, these three equations give 3 - (sin 2 x + sin 2 y + sin 2 z) = 2, whence the desired relation. PROBLEMS 57 3. Let p, q, r and />', q' 9 r' be angles which two straight lines OP and OP' make with Cartesian coordinate axes (Fig. 21). Prove that the angle x between these two directions is given by cos x = cos p cos p' + cos q cos q' + cos r cos r'. FIG. 21. Angle between two directions (perspective drawing). Consider a sphere drawn with the origin as center. Let the radius of that sphere be taken as the unit of length. Produce the arcs ZP and ZP' to their intersections, Q and Q', with XY. Let XQ = y, XQ' = y'. In the triangle PZP', cos x cos r cos r' + sin r sin r' cos (y r y), or cos x = cos r cos r' + sin r sin r' cos y' cos y + sin r sin r' sin ?/' sin y. In the triangle XPQ: cos p = sin r cos y. 58 VII. PROBLEMS In the triangle XFQ': cos p f = sin r' cos y f . Likewise, in the triangles YPQ and YP'Q': cos q = sin r sin y, cos q' sin r' sin y'. Hence cos x = cos r cos r' + cos p cos p' + cos q cos #'. 4. In a triangle ABC (Fig. 22), an arc x, drawn through one of the vertices C, determines on the opposite side c the arcs BD = p and AD = q. Prove that x is given by sin c cos x = sin p cos 6 + sin q cos a. c FIG. 22. Arc through one of the vertices. Let y designate the angle CDB. We have cos a = cos p cos x + sin p sin x cos y, cos b = cos <? cos x sin <? sin x cos y. Multiply the first equation by sin q, the second by sin p, and add the two together, so as to eliminate y. We get sin q cos a + sin p cos b = sin (p + q) cos x, which is the desired relation; and which may also be written PROBLEMS 59 in the form sin p cos 6 + sin q cos a cos x = r^ - . sine Applications. (i) A median in terms of the sides. In this case, p q = ^c. The general formula becomes . c , . , x a + b a b sin jr (cos a + cos 6) cos ^ cos ~ <u L L COS X = = c sm c cos -= (ii) A bisector in terms of the sides. In this case, angle ACD = angle BCD = ^C. We have sin p __ sin a , sin q _ sin b sin C sin y sin C sin i/ ' whence sin p _ sin a sin <? sin 6 ' Now, since p + q = c, sin(c q) _ sin a sin q ~~ sin 6 ' which leads to , sin a + sin b cos c cot q = : r -. . sin 6 sm c Hence 1 vsin b sin c sin q Vl + cot 2 q Vsin 2 a + sin 2 6 + 2 sin a sin 6 cos c Likewise (making q into p and interchanging a and 6), sin a sin c sin Vsin 2 a + sin 2 6 -f 2 sin a sin 6 cos c 60 VII. PROBLEMS On substitution, the general formula becomes sin (a + b) cos x Vsin 2 a + sin 2 6 + 2 sin a sin b cos c (iii) An altitude in terms of the sides. In this case, y = 180 - y = 90. The general formula, thanks to the formulae of right-angled triangles cos a = cos x cos p and cos b = cos x cos q, can be simplified as follows cos a cos b cos x = = . cos p cos q From the relation cos p _ cos a cos q cos b ' by transformations similar to those used in (ii), we deduce , cos a cos b cos c tan q = j -7 , cos 6 sin c whence 1 __ /. , , _ Vcos 2 a + cos 2 6 2 cos a cos 6 cos c ^ J. | VlAlli U _ . cos q cos o sin c It follows that 1 cos x = - Vcos 2 a + cos 2 6 2 cos a cos b cos c. sin c 5. A point P is located (Fig. 23) between two great circles AC and BC. The distances from P to these great circles are p and g, respectively. The angle ACB between the great circles is C. What is the distance x from P to the intersec- tion of the two given great circles? Let ACP = y, whence BCP = C - y. The two right- angled triangles give sin p sin x sin y, sin q = sin x sin (C y). PROBLEMS Eliminate y. The first equation gives sm p , sin y = -T - , whence cos y = Substituting in the second equation, we get 61 x """ sn sin q = sin C Vsin 2 x sin 2 p cos C sin p. c FIG. 23. Point between two great circles. Transposing and squaring, . , . , (sin q + cos C sin t>) 2 sin' x - sm* p = ^^ , whence sin 2 x sin 2 p sin 2 C + sin 2 q + cos 2 C sin 2 p + 2 sin p sin g cos C sin 2 C and, finally, Sm X ~ s i n c Vsin 2 p + sin 2 5 + 2 sin p sin q cos C. Numerical Application. What is the latitude of a point in the Northern hemisphere, knowing its distances from the zero meridian (45) and from the 90-meridian (30)? 62 VII. PROBLEMS Let L be the latitude. The above formula becomes cos L = Vsin 2 45 + sin 2 30, T ^ cos L = -- , whence L = 30. Remark. Note that, in this particular example, the point is located within a tri-rectangular triangle, so that (Problem 2) sin 2 45 + sin 2 30 + sin 2 L = 1. 6. In a spherical triangle (Fig. 24), the cosine of the arc which connects the middle points of two sides is proportional to the cosine of half the third side ; the factor of proportionality is the cosine of one half of the spherical excess. Required to prove: *)' h cos x = cos E cos - , cos y = cos E cos ^ > cos z - cos E cos - . FIG. 24. Arc that connects the mid- points of two sides. Proof. In the triangle AEF, b c , . b . c , cos x = cos ~ cos 5 + sin ^ sm ^ cos A. PROBLEMS 63 In the triangle ABC, cos a cos b cos c + sin b sin c cos A. Taking cos A from each equation, and equating the two values, we get b c COS X COS COS jr , 2 2 cos a cos b cos c . b . c . . b . c b c' sm - sm x 4 sin ^ sin ^ cos ^ cos ~ L Jj A Hence cos a cosx = A C 4 COS jr COS jr 6 c cos a + 2 cos 2 o + 2 cos 2 X 4 COS pr COS ^ 2i tli CQS a + (1 + cos 6) + (1 + cos c) ^ 5 c 4 cos - cos ^ z ^ 1 -f cos a + cos 6 + cos c - T - . . 6 c 4 cos ~ cos ^ This expression, divided by cos -= , is equal to cos E (Euler's formula) . Hence cos x = cos E cos ~ . Likewise for the other two formulae. 64 VII. PROBLEMS *Second Proof. In a plane triangle c' = a' cos B' + b' cos A'. This formula, applied to the derived triangle (Fig. 11), where A' = 180 - A, B' = E, C' = A - E, gives be a ^ . b . c . cos 55 cos K = cos ~ cos E sm ^ sm ~ cos A 2t & Z & or ^ a b c , . b . c . cos E cos n = cos o cos 5 + sm - sm ^ cos A, A A i t which is equal to cos x, as seen in the triangle AEF (Fig. 24). Remark. The formula cos x = cos E cos ~ & can be derived directly in the following way. o FIG. 25. Geometrical construction (perspective drawing). Construction. Join (Fig. 25) the vertices A, B, and C to the center of the sphere of unit radius (OA = OB = OC = 1). PROBLEMS 65 Join to the middle points E and F of the sides b and c. Draw the chords AB, BC, and CA. Join H, the point of intersection of OE and CA, to K, the point of intersection of OF and AB. Proof. The formula giving a relation between the three sides and one of the angles of a plane triangle is applied to the triangle OHK. Since the angle COA = 6, we have, in the plane OCA, Likewise OH = cos ~ . / OK = COS j: . Since E is the middle point of the arc CA, H is the middle point of the chord CA. Likewise, K is the middle point of C*H\ n. AB. Hence, in the triangle ABC, HK = ^ = sin ~ , for A & the chord subtending an arc is equal to twice the sine of half the arc ( CB = 2 sin | j . Now the angle EOF is measured by the arc EF = x. By plane trigonometry, we have, in the triangle OHK, sin 2 jr = cos 2 ~ + cos 2 ~ 2 cos ^ cos ~ cos #, A U Li & 4 whence 2 cos 2 ~ + 2 cos 2 | - 2 sin 2 ~ cos x = z , A b c 4 cos cos ^ & A (1 + cos 6) + (1 + cos c) (1 cos a) a cos x = -^ ^- i -^ L T cos x , . a b c a 2 ' 4 COS <r COS ^ COS ^ COS ^ COS X = COS E COS x . 66 VII. PROBLEMS 7. The area of an equilateral triangle is S/n, where S desig- nates the area of the sphere. Find the side x of the triangle. Let A be the angle of the triangle. By Girard's theorem (Appendix I), 1 }(3A - 180) , . n + 4 - no _ - * L whence A = 60 . n 360 * n The side x is given in terms of the angles by the fundamental formula (10) cos A = cos 2 A + sin 2 A cos x, from which we deduce cos A cos A cos x 1 cos A . 9 A 2 sm* or, after substitution, COS X = Numerical Application. Take n = 4. Show that x *arc cos ( -J). cos 120 -\ , x = 180 y, where cos y = J. log cosy = 1.522 8787 9002 215 178 5 36 5 y = 7031'43.6" x = 109 28'16.4". PROBLEMS 67 8. Given the dihedral angle (or edge angle) a between two slant faces of a regular n-sided pyramid, calculate the face angle x between two adjacent slant edges (Fig. 26). s Fio. 26. A regular pyramid. The base of the pyramid is an n-sided regular polygon; the interior angle between two sides is equal to y = ^-? 180. n Around the vertex S draw the unit sphere, so as to replace the trihedron S by a spherical triangle. Decomposing this isosceles triangle in two right-angled triangles, we get im- mediately whence . y . ot x sm ^ = sm ^ cos ~ > sin Q()0 x n COS^r 2 . a sm- 68 VII. PROBLEMS x Discussion. The problem is possible only if < cos ~ < 1- 4U X X It is obvious that cos cannot be negative, as must be ) smaller than 90. ) The first condition demands sin 90 > 0, n ' which implies n >: 3. ( Note that sin -= is always positive, j The second conditions requires sin 75 > sin 90, A n that is to say a > 180, which is evident (by geometry). Numerical Application. In a regular hexagonal pyramid, let a = 170. The general formula becomes x sin 60 1 .937 5306 COS 2 sin 85 1.998 3442 log cos I = 1.939 1864 1953 89 84 5 I = 2937'7.4" x = 5914'14.8". 9. Volume of a parallelepiped. Let OA = a, OB = 6, OC = c, be three edges making with one another (Fig. 27) the angles BOG = a, COA = ft AOB = 7. Through C pass a plane normal to OA and con- taining CD, perpendicular to the plane AOB. The angle PROBLEMS 69 CED = A is the dihedral angle of the edge OA. Draw the unit sphere around the vertex 0. FIG. 27. Volume of a parallelepiped. V = ab sin y . CD = ab sin 7 . CE sin A = abc sin 7 . sin /3 sin A. But A A sin A = 2 sin cos = 2 Vsin p sin (p a) sin (p ff) sin (p 7) sin /3 sin 7 ' in which 2p = a + + 7, from the formulae (8). It follows that V = 2a6c Vsin p sin (p a) sin (p 0) sin (p 7). Special Cases. Volume of a rhombohedron (b = c = a, /? = 7 = a) : TT- r 7 V = 2a 3 . sin 3 a -= . Volume of a right-angled parallelepiped (a = = 7 = 90) : V = a6c. Volume of a cube (6 = c = a, a = j3 = 7 = 90) : V = a 3 . CHAPTER VIII EXERCISES 1 1. The face angles of a trihedron measure 50. Calculate one of the dihedral angles (edge angles) . 2. In a triangle ABC, given b = 45, c = 60, A = 30. Find B. 3. Two triangles ABC and BCD, both right-angled at C, have a side of the right angle in common, BC = 5936'30". The other side about the right angle is known in each case: CA = 6422', CD 52 5'. Find the hypotenuses AB and BD. 4. Given, in a right-angled triangle (C = 90), cose = 1/V3 and b = 45. Calculate the other two angles, each from the data only, and express the area of the triangle as a fraction of the area S of the sphere on which it is drawn. 5. Show that if (C - E) : A : (B - E) = 2 : 3 : 4, then 2 cos i(C - E) sin %(b + c)/sin \a. [Hint: it is known that if the angles of a plane triangle A'B'C' are in the ratios A' : B' : C' = 2 : 3 : 4, then 2 cos JA' = (a' + c')/b'.] 6. An equilateral triangle ABC, with a = b = c = 30, is drawn on a sphere of radius R. Find the area of the triangle expressed in dm 2 if R = 6w. 7. Let O, O', O" be the middle points of the three edges of a cube intersecting in the same vertex C. Calculate the elements of the trihedron whose edges are OC, 00', OO". 8. In an isosceles triangle, given b = c = 32 and B = 78. Find the side a. 9. In a regular tetragonal pyramid, the altitude is equal to one half the diagonal of the base. Calculate the dihedral angle between two adjacent slant faces. 10. Express sin (a 4- b) /sin c in terms of the three angles of an oblique-angled triangle. 1 Most of these exercises are questions that were asked by Cesaro at the entrance examination of the University of Lie"ge. 70 EXERCISES 71 11. The arc x that connects the middle points of two sides of an equilateral triangle of side a is given by 2 sin \x tan Ja. (Compare Problem 6, Chapter VII.) 12. In a regular tetragonal pyramid, the angle of a face at the apex measures 60. Calculate the dihedral angle between two ad- jacent slant faces. 13. In a regular pentagonal pyramid, the angle of a face at the apex measures 30. Calculate the dihedral angle between two ad- jacent slant faces. 14. The altitude CH of a triangle ABC, right-angled at C, inter- sects the hypotenuse AB at H; arc AH = b' and arc BH = a'. Prove that *t "' sin 2 b' _ I . n T ~ 1 . sin 2 a sin 2 6 15. Given a tri-rectangular trihedron O,XYZ. Plot OP = 2, OQ = 3, OR = 4, on the edges OX, OY, OZ, respectively. Calcu- late the elements of the trihedron P, OQR. 16. Let ipi> PI be the longitude and polar distance (or colatitude) of a point PI; <?A, PA, those of a point A. A 180-rotation about A brings PI in P2. What are the longitude <pt and the polar distance P2 of P 2 ? 17. Let AB - p, AC = q, AD = r be three edges of a right- angled parallelepiped . Calculate the elements of the trihedron whose edges are DA, DB, DC. Numerical application: p = 3, q = 4, r = 5. 18. In a quadrantal triangle (c = 90), given C and 2p. Find (1) sin a sin 6; (2) sin A sin B. 19. In a right-angled triangle (C = 90), given the hypotenuse c and the spherical excess 2E. Find: (1) sin a sin 6; (2) sin A sin B. 20. Given an equilateral triangle whose side is equal to 60. Find the side of the triangle formed by erecting, at each vertex, a perpen- dicular to the corresponding bisector. 21. Solve a quadrantal triangle (a = 90), given: (1) c, C; (2) A, b + c. 22. Prove the formulae of a quadrantal triangle (a = 90) : sin p SB cos JB cos iC/sin iA, cos p = sin JB sin iC/sin JA, tan p = cot iB cot JC. 72 VIII. EXERCISES 23. Consider a regular pentagonal prism with altitude h and the side of the base equal to a. Join 0, one of the base vertices, to P and Q, the top vertices opposite the two adjacent vertical faces. Calculate the angle POQ. Numerical application: a = 4, h = 8. 24. Same question for a hexagonal prism. Numerical application : a = 5, h - 5. 25. Join a point P, inside an oblique-angled triangle ABC, to the vertices and produce the arcs AP, BP, CP to their intersections, D, E, F, with the sides of ABC. Prove that sinCE sin AF sin BD sin EA sin FB sin DC 26. In a tri-rectangular triangle ABC, join the middle points B' and C' of the sides AB and AC by the arc B'C'. Find the ratio of the areas of the triangles AB'C' and ABC. 27. In a triangle ABC, let a, 0, 7 be the lengths of the bisectors. Prove: (1) cot a cos JA -f cot /3 cos JB -f- cot 7 cos JC = cot a -f- cot b -f- cot c; cot a cot j8 (2) cos JA(cos B H- cos C) cos JB(cos C -f- cos A) cot 7 cos iC(cos A -f cos B) ' . 2 Vsin b sin c sin p sin (p a) Vsin 2 6 + sin 2 c -f- 2 sin 6 sin c cos a 28. In a triangle ABC, given a - b 90 and C 60. Join any point P inside the triangle to the vertices A, B, C by arcs x, y, z. Find a relation between x, y y z, 29. In a quadrilateral ABCD, given :B C = D = 90, BD = a, and AB b. Calculate the diagonal AC. 30. Show that the radii R c , R, Ra of the circumscribing, inscribed, and escribed circles of a spherical triangle are given respectively by: 2 P - cosS tan KC where 2S = A + B -f C; EXERCISES 73 sm(p a) sin(p 6) sin(p c) where 2p a 4- b + c; tan 2 R % sin p tan 2 R = sin p sin(p - 6) sin(p c) sin(p a) ' where R is the radius of the escribed circle which touches a. ANSWERS TO EXERCISES Numerical applications have been calculated with five-place tables. The answer is followed, between parentheses, by the number of the formula to be applied. 1. 6658' (8). 2. B = 496'24" (12). 3. AB = 7721'28", BD = 7153'14" (13). 4. A = 60, B = 45, T = S/48 (15 and 16, no tables of logarithms needed). 5. The desired relation is read off the triangle of elements (Fig. 9). 6. E = 3.5, T = 439.8 dm 2 (5a and Appendix 1). 7. COO' = COO" = 45, O'OO" = 60, edge angle OC = 90, edge angle OO' = edge angle OO" = 5444' (13 and 18). 8. a = 1448'17" (16). 9. 10928' (14). 10. sin(a -f b)/sin c = (cos A -f cos B)/(l - cos C). Apply (3) (4). 11. Apply (5a). 12. 10928' (14). 13. 11346' (17). 14. Apply (15) twice. Use sin z z + cos 2 a; = 1. 15. Edge angle PO = 90, OPR = arctan 2 = 6326' 5", OPQ - arctan (3/2) = 5618'35", edge angle PR = 59 11 '33", edge angle PQ = 6724 / 41", RPQ - 7538'11". Apply (13) and (18), the latter twice. 16. cos p2 = cos 2pA cos pi + sin 2pA sin pi cos (V?A ^0, sin (<f>2 <PA) sin pi sin (V>A ^0/sin p 2 . Apply Problem 4, Ch. VII and (7). (This question is encountered in crystallography.) 17. The spherical triangle ABC is right-angled at A. Its parts are: a = 4757 / 51 // , b = 3839 r 36", c = 3057'50 // , B = 5715 / 15", C 4350'17 /; . Apply (13) and (18). First, determine 6 and c by plane trigonometry. 18. sin a sin b = sin 2p/(l + cos C), sin A sin B = sin 2p (1 cos C). Apply (8b), then combine with (9). 74 ANSWERS TO EXERCISES 75 19. sin a sin b = sin2E(l + cos c), sin A sin B = sin2E/(l cose). Apply (11) and (8), or consider polar triangle of quadrantal triangle in preceding exercise. 20. 90 (8a, then 16; no tables of logarithms needed). 21. Given c, C. Write sin & = cosc/cosC, sin A = sinC/sinc, sin B = cot c/cot C. Apply (7) (9) (12). Given A, b + c. Express cos i(B C) and cos i(B -f C) in terms of J(& -f c) and JA by means of (3) (4), hence B and C. Like- wise express sin i(& c) in terms of $(B C) and JA by means of (3) or (4). Since (6 -\- c) is known, you get 6 and c. 22. Express the corresponding relations in the (right-angled) polar triangle of the given triangle. Apply the Law of Sines to Fig. 11. 23. sin i(POQ) = sin 18 sm c, with tan c 2(a/h) cos 36. Ap- ply (15). For a = 4 and h = 8, POQ = 2224'53". 24. Apply (7) or (15). For a/h = 1, POQ = 5119' 4". 25. Apply (9). Note three pairs of equal angles having common vertex P. (This question is encountered in crystallography.) 26. 0.216 (14 or 7, and Girard's Theorem, Appendix 1). 27. (1) Apply (12). (2) Apply (10) (9). (3) Apply (12) twice. 28. 4 (cos 2 x -f cos 2 y cos x cos y) = 3 sin 2 z. Apply (7) and sin 2 a -f cos 2 a = 1. 29. Let x = AC. It is given by the biquadratic equation cos 4 x cos 2 a cos 2 b cos 2 x cos 2 a cos 2 b sin 2 6 = 0. Apply (13) three times and (15) twice. 30. Consider a circle inscribed in a spherical triangle. It is tangent to the sides, The spherical distances from a vertex to the nearest two points of tangency are equal. The radii to the points of tangency are perpendicular to the corresponding sides. Apply the formulae of right-angled triangles. APPENDIX 1 SPHERICAL AREAS 1. Sphere. The area S of a sphere is equal to four times that of a great circle, that is to say S = 4irR 2 , if R designates the radius. 2. Lune. The area of a lune of angle A is to the area of the sphere as A is to 360. Consider a number of meridians; they intersect in the North and South Poles. Any two meridians bound a lune (for example, the meridians 75W and 80W). The angle of the lune is the angle between the two meridians (in this case, 5); it is measured by the arc they intercept on the equator. Obviously lunes of equal angles have equal areas (for they can be made to coincide). The areas of lunes are proportional to their angles (this is proved in the same manner as the proportionality of angles at the center of a circle to the arcs they subtend), whence the desired proposition. 3. Spherical Triangle. GIRARD'S THEOREM: The area of a spher- ical triangle is to the area of the sphere as half the spherical excess (expressed in degrees) is to 360. FIG. 28. Area of a spherical triangle. Consider a triangle ABC. Join its vertices A, B, C to the center O of the sphere (Fig. 28) and produce to the antipodes A', B' f C'. Complete the great circles that form the sides of the triangle ABC. 76 SPHERICAL AREAS 77 The hemisphere in front of the great circle BCB'C' is thus divided into four triangles: Ti ABC, T 2 AB'C', T, - AB'C, T 4 = ABC'. From the construction it follows that the triangle A'BC is equal to the triangle T 2 =* AB'C'. Hence, designating by lune A the area of a lune of angle A, Ti -f T 2 = lune A. Moreover Ti + T, = lune B and Ti + T 4 = lune C. On adding and transposing, 2 Ti = lune A -f lune B + lune C - (Ti + T 2 -f T 3 -f T 4 ), whence 2Ti A + B + C - 180 ^ 2JS 8 360 " 360 and, finally, II JL S 360* APPENDIX 2 FORMULAE OF PLANE TRIGONOMETRY DEFINITIONS: versine, coversine, exsecant, coexsecant, haversine. vers x = 1 cos x, exsec x = sec x 1 , co verso: = 1 sin x, coexsecx = cscx 1, hav x = J vers x = i(l cos x) = sin 2 Jx. sin 30 = i, sin 45 = ^ , sin 60 = ^? , sin 90 = 1. /L i Complementary angles, x and (90 x), have complementary trigonometric functions, with the same sign: sin x = cos (90 - x), tan x = cot (90 - x), sec x = esc (90 x). Supplementary angles, x and (180 x), have the same trigo- nometric functions, with opposite signs except sine and cosecant. Antisupplementary angles, x and (180 -f x), have the same trigo- nometric functions, with opposite signs except tangent and cotangent. Equal angles of opposite signs, x and x, have the same trigo- nometric functions, with opposite signs except cosine and secant. 1 cot x cosx = Vl + cot 2 x ' sin (x y) = sin x cos y db sin y cos x, cos (x db y) = cos x cos y =F sin x sin #. , N tan x . tan y sin 2x = 2 sin x cos x, 2 sin 2 Jx = 1 cos x, cos 2x = cos 2 x sin 2 x, 2 cos 2 Jx = 1 + cos x, , 2 tan x x o i 1 cos x tan 2x = ; - r r , tan 2 \x = ^ ; - . 1 - tan 2 x ' 3 1 4- cos a? 78 FORMULAE OF PLANE TRIGONOMETRY 79 1 -f- tan x ,AKO , \ cot as -t- 1 , //4 _ x - IT = tan (45 H- x), ? = cot (43 x). 1 tan a; v " cot x 1 v ' 2 sin x cos y = sin (x -f- y) -f- sin (x y), 2 cos x cos 2/ cos (x -f- ?/) -f- cos (x y}, 2 sin x sin ?/ = cos (x y) cos (x + 2/) sin P + sin Q = 2 sin J(P -f- Q) cos i(P - Q), sin P - sin Q = 2 cos ^(P + Q) sin J(P - Q), cos P -f cos Q = 2 cos i(P -f Q) cos J(P - Q), cos P - cos Q = 2 sin $(P + Q) sin KQ - P). cos P cos Q . . /T ^ , ,^.1. . , /rA -TJV - tan i(P + Q) tan i(Q - P) ' SOLUTION OF TRIANGLES a b c LAW OF SINES: sin A sin B sin C " c 6 cos A Corollary: cot V = LAW OP COSINES: o 2 = b 2 + c 2 26c cos A. T _, tan KB - C) tanKB-C) b - c LAW OP TANGENTS: tan >(B + c) - cot iA = 5^^ Half -angle formulae: - (2p = a -f- 6 + c) cos 2 - = p(p r "" a) , 2 be A (p-6)(p -c) TT = - 7 - r - . 2 p(p a) Area o/^e triangle: T = Vp(p a)(p 6)(p c). INDEX Altitude 60 Ambiguity 41 Analogies 22 Angle between two direc- tions 57 Angular distance 5, 51 Antiparallel 8 Antisupplementary angles . 78 Arc through one of the ver- tices 58 Area of a lune 76 Area of a sphere 76 Area of a spherical triangle 76 Astronomy 6 B Bisector 59 Cagnoli's formula 25 Calculations 41, 44 Cesaro 70 Cesaro's key-triangles 13 Checking calculations 45 Circumscribing circle 72 Coexsecant 78 Complementary angles. ... 78 Concyclic points 15 Consecutive parts 32 "Cot-sin-cos" formula 34 Coversine 78 Crystallography 5, 74, 75 Cube 69 Cyclic parts 40 Delambre's formulae 22 "Derived triangle" 16 Dihedral angle 3 Equal angles of opposite signs 78 Escribed circle 72 "Eulerian" spherical tri- angle 4 Euler's formula 23 Examples of calculations . . 45 Exsecant 78 F Fundamental formulae. ... 41 Gauss 22 Geodesy 5 Geographical problem 51 "Geoid" 6 Geometry 5 Girard's theorem 66, 76 Great circle 3 H Half-angles 29 Half-sides 30 Haversine 78 81 82 INDEX Hypotenuse 36 Hypotenuse and two sides or two angles 37 Hypotenuse, one side, and one angle 37 Inscribed circle 72 Point between two great circles 61 Polar triangles 18 Poles 18 Polygon 3 Precision 52 Problems (with solutions) . . 55 Product formulae 37 Pyramid 67 Law of cosines 79 Law of sines 79 Law of tangents 79 Lhuilier's formula 24 Logarithmic computation, formulae adapted to 42 Lune 16, 76 M Median 59 Mid-points of two sides ... 62 N Napier's analogies 22 Napier's rule 39 Navigation 6 Negative characteristic 44 O Oblique-angled spherical triangles 28, 41 Parallelepiped 68 Perimeter 4 Plane triangle 20 Plane trigonometry formu- lae 78 Quadrantal triangles 41 Ratios of cosines and co- tangents 38 Ratios of sines and tangents 37 Rhombohedron 69 Right-angled parallelepiped 69 Right-angled triangles. 36, 41 Right-sided triangles 41 Sine of the trihedral angle . 21 Solution of triangles 41 Spherical areas 76 Spherical distance 5, 51 Spherical excess 14, 23, 24, 25, 26 Spherical polygon 3 Spherical triangle 4, 21 Stereographic projection . . 7 Sub-contrary 8 Supplementary angles 78 Surveying 5 Three angles and one side . 30 Three sides and one angle . 28 INDEX 83 "Triangle of elements". ... 13 Two sides and opposite Trihedra, complementary . 16 angles 29 Trihedron 4 Two sides, their included Tri -rectangular triangle. . 55, angle, and the angle op- 56, 62 posite one of them 32 Two angles and one side, or two sides and one angle (in a right-angled tri- Versine 78 angle) 38